The last attribute of glVertexAttribPointer is of type const GLvoid*. But is it really a pointer? It is actually an offset. If I put 0, it means an offset of 0 and not a null pointer to an offset. In my engine, I use this function:
void AbstractVertexData::vertexAttribPtr(int layout) const
{
glVertexAttribPointer(layout,
getShaderAttribs()[layout]->nbComponents,
static_cast<GLenum>(getShaderAttribs()[layout]->attribDataType),
getShaderAttribs()[layout]->shouldNormalize,
getVertexStride(layout),
reinterpret_cast<const void*>(getVertexAttribStart(layout)));
}
getVertexAttribStart returns an intptr_t. When I run drmemory, it says "uninitialized read" and I want to remove that warning. This warning comes from the reinterpret_cast. I can't static_cast to a const void* since my value isn't a pointer. What should I do to fix this warning?
Originally, back in OpenGL-1.1 when vertex arrays got introduces, functions like glVertexPointer, glTexCoordPointer and so on were accepting pointers into client address space. When shaders got introduced they came with arbitrary vertex attributes and the function glVertexAttribPointer follows the same semantics (this was in OpenGL-2.1).
The buffer objects API was then reusing existing functions, where you'd pass an integer for a pointer parameter.
OpenGL-3.3 core eventually made the use of buffer objects mandatory and ever since the glVertexAttribPointer functions being defines with a void* in their function signature are a sore spot; I've written in extent about it in https://stackoverflow.com/a/8284829/524368 (but make sure to read the rest of the answers as well).
Eventually new functions got introduced that allow for a more fine grained control over how vertex attributes are accessed, replacing glVertexAttribPointer, and those operate purely on offsets.
Related
I want to find some ACSL annotation that can be applied to a function or function pointer to indicate that it has the property of referential transparency. Some way to say "this function will always return the same value when given the same arguments". So far I haven't found any such way. Can anyone point me to a way to express that?
Maybe some way to refer to an arbitrary logic function? If I could name an unknown logic boolean uknown_function(void* a, void* b) = /* this is unkown */; then I could document a function as having a postcondition that it's \result is equal to this arbitrary/unknown logic function?
The larger context is trying to do type-erased comparisons. I want to generally express the concept of "the user has given me void*s to work with and a bool (*)(void const*, void const*) to compare them with, and the user is guaranteeing to me that the function provided really is a strict partial order over whatever those pointers point to." If I had that, then I could start to describe properties of these type-erased objects being sorted, for example.
There is indeed no direct possibility to do that in ACSL: a function contract only specifies what happens during a single call of the function. You could indeed rely on a declared but left undefined logic function, with a reads clause that specifies the part of the C memory state that the function will need to compute its result, e.g.
/*# logic boolean unknown_function{L}(int* a, int* b) reads a[0 .. 1], b[2 .. 3]; */
but if you work with void *, without knowing the size of the underlying objects, this might be tricky to specify: unless the result of unknown_function relies solely on the value of the pointer, and not the content of the pointed object, in which case you don't need that reads trick.
Note in addition that contracts over function pointers are not supported yet, which will probably be an issue for what you intend to do if I understand correctly your last paragraph.
Finally, you might be interested in an upcoming plug-in, RPP, that proposes a way to specify, prove, and use properties relating several calls of one or more C function(s). It is described here and here, and a public release should happen in a not-too-distant future.
If I have the following code:
// objective C++ code .mm
id<MTLTexture> texture = ...;
void* ptr = (void*)CFBridgingRetain(texture);
share_ptr_with_native_code(ptr);
[texture do_stuff]; // is this valid?
// native code .cpp
void share_ptr_with_native(void* ptr)
{
ptr->do_stuff();
CFBridgingRelease(ptr);
}
Will texture be valid and retained by ARC again after the call to share_ptr_with_native()?
Other than various errors in your code snippet, yes, the line in question is valid. ARC continues to maintain its own strong reference to object while it's still in use in the top code, in addition to the one that you become responsible for. CFBridgingRetain() has a +1 effect on the retain count of the object, hence "retain" in its name.
Even everything said is right, it would be nicer if you change your
CFBridgingRelease(ptr);
to
CFRelease(ptr) .
__bridge_retained or CFBridgingRetain casts an Objective-C pointer to a Core Foundation pointer and also transfers ownership to you.
You are responsible for calling CFRelease or a related function to relinquish ownership of the object.
Taken from https://developer.apple.com/library/content/documentation/CoreFoundation/Conceptual/CFDesignConcepts/Articles/tollFreeBridgedTypes.html.
Isn't a unique_ptr essentially the same as a direct instance of the object? I mean, there are a few differences with dynamic inheritance, and performance, but is that all unique_ptr does?
Consider this code to see what I mean. Isn't this:
#include <iostream>
#include <memory>
using namespace std;
void print(int a) {
cout << a << "\n";
}
int main()
{
unique_ptr<int> a(new int);
print(*a);
return 0;
}
Almost exactly the same as this:
#include <iostream>
#include <memory>
using namespace std;
void print(int a) {
cout << a << "\n";
}
int main()
{
int a;
print(a);
return 0;
}
Or am I misunderstanding what unique_ptr should be used for?
In addition to cases mentioned by Chris Pitman, one more case you will want to use std::unique_ptr is if you instantiate sufficiently large objects, then it makes sense to do it in the heap, rather than on a stack. The stack size is not unlimited and sooner or later you might run into stack overflow. That is where std::unique_ptr would be useful.
The purpose of std::unique_ptr is to provide automatic and exception-safe deallocation of dynamically allocated memory (unlike a raw pointer that must be explicitly deleted in order to be freed and that is easy to inadvertently not get freed in the case of interleaved exceptions).
Your question, though, is more about the value of pointers in general than about std::unique_ptr specifically. For simple builtin types like int, there generally is very little reason to use a pointer rather than simply passing or storing the object by value. However, there are three cases where pointers are necessary or useful:
Representing a separate "not set" or "invalid" value.
Allowing modification.
Allowing for different polymorphic runtime types.
Invalid or not set
A pointer supports an additional nullptr value indicating that the pointer has not been set. For example, if you want to support all values of a given type (e.g. the entire range of integers) but also represent the notion that the user never input a value in the interface, that would be a case for using a std::unique_ptr<int>, because you could get whether the pointer is null or not as a way of indicating whether it was set (without having to throw away a valid value of integer just to use that specific value as an invalid, "sentinel" value denoting that it wasn't set).
Allowing modification
This can also be accomplished with references rather than pointers, but pointers are one way of doing this. If you use a regular value, then you are dealing with a copy of the original, and any modifications only affect that copy. If you use a pointer or a reference, you can make your modifications seen to the owner of the original instance. With a unique pointer, you can additionally be assured that no one else has a copy, so it is safe to modify without locking.
Polymorphic types
This can likewise be done with references, not just with pointers, but there are cases where due to semantics of ownership or allocation, you would want to use a pointer to do this... When it comes to user-defined types, it is possible to create a hierarchical "inheritance" relationship. If you want your code to operate on all variations of a given type, then you would need to use a pointer or reference to the base type. A common reason to use std::unique_ptr<> for something like this would be if the object is constructed through a factory where the class you are defining maintains ownership of the constructed object. For example:
class Airline {
public:
Airline(const AirplaneFactory& factory);
// ...
private:
// ...
void AddAirplaneToInventory();
// Can create many different type of airplanes, such as
// a Boeing747 or an Airbus320
const AirplaneFactory& airplane_factory_;
std::vector<std::unique_ptr<Airplane>> airplanes_;
};
// ...
void Airline::AddAirplaneToInventory() {
airplanes_.push_back(airplane_factory_.Create());
}
As you mentioned, virtual classes are one use case. Beyond that, here are two others:
Optional instances of objects. My class may delay instantiating an instance of the object. To do so, I need to use memory allocation but still want the benefits of RAII.
Integrating with C libraries or other libraries that love returning naked pointers. For example, OpenSSL returns pointers from many (poorly documented) methods, some of which you need to cleanup. Having a non-copyable pointer container is perfect for this case, since I can protect it as soon as it is returned.
A unique_ptr functions the same as a normal pointer except that you do not have to remember to free it (in fact it is simply a wrapper around a pointer). After you allocate the memory, you do not have to afterwards call delete on the pointer since the destructor on unique_ptr takes care of this for you.
Two things come to my mind:
You can use it as a generic exception-safe RAII wrapper. Any resource that has a "close" function can be wrapped with unique_ptr easily by using a custom deleter.
There are also times you might have to move a pointer around without knowing its lifetime explicitly. If the only constraint you know is uniqueness, then unique_ptr is an easy solution. You could almost always do manual memory management also in that case, but it is not automatically exception safe and you could forget to delete. Or the position you have to delete in your code could change. The unique_ptr solution could easily be more maintainable.
If you capture a strong reference to self under ARC in an objective-C style block, you need to use a __weak pointer to avoid an ARC "retain cycle" problem.
// Right way:
- (void)configureBlock {
XYZBlockKeeper * __weak weakSelf = self;
self.block = ^{
[weakSelf doSomething]; // capture the weak reference
// to avoid the reference cycle
}
}
I really don't know what a retain cycle is, but this answer describes it a bit. I just know you should use a __weak pointer for Objective-C style blocks. See Avoid Strong Reference Cycles when Capturing self.
But my question is, do I need to create a weak pointer when capturing self under a C++ <functional> block?
- (void)configureBlock {
self.block = [self](){
[self doSomething]; // is this ok? It's not an objective C block.
}
}
C++ lambdas can captured variables either by value or by reference (you choose when you declare the lambda how to capture each variable).
Capturing by reference is not interesting, because references to local variables become invalid after you leave the variable's scope anyway, so there is no memory management issues at all.
Capturing by value: if the captured variable is an Objective-C object pointer type, then it gets interesting. If you are using MRC, nothing happens. If you are using ARC, then yes, the lambda "retains" captured variables of object pointer type, as long as they are __strong (not __weak or __unsafe_unretained). So, yes, it would create a retain cycle.
#include<stdio.h>
int main()
{
const int sum=100;
int *p=(int *)∑
*p=101;
printf("%d, %d",*p,sum);
return 0;
}
/*
output
101, 101
*/
p points to a constant integer variable, then why/how does *p manage to change the value of sum?
It's undefined behavior - it's a bug in the code. The fact that the code 'appears to work' is meaningless. The compiler is allowed to make it so your program crashes, or it's allowed to let the program do something nonsensical (such as change the value of something that's supposed to be const). Or do something else altogether. It's meaningless to 'reason' about the behavior, since there is no requirement on the behavior.
Note that if the code is compiled as C++ you'll get an error since C++ won't implicitly cast away const. Hopefully, even when compiled as C you'll get a warning.
p contains the memory address of the variable sum. The syntax *p means the actual value of sum.
When you say
*p=101
you're saying: go to the address p (which is the address where the variable sum is stored) and change the value there. So you're actually changing sum.
You can see const as a compile-time flag that tells the compiler "I shouldn't modify this variable, tell me if I do." It does not enforce anything on whether you can actually modify the variable or not.
And since you are modifying that variable through a non-const pointer, the compiler is indeed going to tell you:
main.c: In function 'main':
main.c:6:16: warning: initialization discards qualifiers from pointer target type
You broke your own promise, the compiler warns you but will let you proceed happily.
The behavior is undefined, which means that it may produce different outcomes on different compiler implementations, architecture, compiler/optimizer/linker options.
For the sake of analysis, here it is:
(Disclaimer: I don't know compilers. This is just a logical guess at how the compiler may choose to handle this situation, from a naive assembly-language debugger perspective.)
When a constant integer is declared, the compiler has the choice of making it addressable or non-addressable.
Addressable means that the integer value will actually occupy a memory location, such that:
The lifetime will be static.
The value might be hard-coded into the binary, or initialized during program startup.
It can be accessed with a pointer.
It can be accessed from any binary code that knows of its address.
It can be placed in either read-only or writable memory section.
For everyday CPUs the non-writeability is enforced by memory management unit (MMU). Messing the MMU is messy impossible from user-space, and it is not worth for a mere const integer value.
Therefore, it will be placed into writable memory section, for simplicity's sake.
If the compiler chooses to place it in non-writable memory, your program will crash (access violation) when it tries to write to the non-writable memory.
Setting aside microcontrollers - you would not have asked this question if you were working on microcontrollers.
Non-addressable means that it does not occupy a memory address. Instead, every code that references the variable (i.e. use the value of that integer) will receive a r-value, as if you did a find-and-replace to change every instance of sum into a literal 100.
In some cases, the compiler cannot make the integer non-addressable: if the compiler knows that you're taking the address of it, then surely the compiler knows that it has to put that value in memory. Your code belongs to this case.
Yet, with some aggressively-optimizing compiler, it is entirely possible to make it non-addressable: the variable could have been eliminated and the printf will be turned into int main() { printf("%s, %s", (b1? "100" : "101"), (b2? "100" : "101")); return 0; } where b1 and b2 will depend on the mood of the compiler.
The compiler will sometimes take a split decision - it might do one of those, or even something entirely different:
Allocate a memory location, but replace every reference with a constant literal. When this happens, a debugger will tell you the value is zero but any code that uses that location will appear to contain a hard-coded value.
Some compiler may be able to detect that the cast causes a undefined behavior and refuse to compile.